Count of all combinations of a set

Question

All right, sorry in advance if this exists on here already. What I'm after is figuring out all possible orders of a set.

So I have 2 items

1,2
2,1
=2

3 items

1,2,3
1,3,2
2,1,3
2,3,1
3,1,2
3,2,1
=6

I can't seem to find a correlation I can burn down to an equation.

Any ideas? Thanks.

Answer 1

For computing the number of permutations of $n$ elements of a set, (distinct arrangements of elements of the set), there are $$n! = n(n-1)(n-2) \cdots 2 \cdot 1$$ possible combinations.

Answer 2

for $\emptyset$ with $0$ elements we have 1 permutation

for $\{1\}$ set with $1$ element we have $$1$$ permutation

for $\{1,2\}$ set with $2$ elements we have $$2=1\cdot2=2!$$ permutations

$12$

$21$

for $\{1,2,3\}$ set with $3$ elements we have $$6=1\cdot2\cdot3=3!$$ permutations

$123,132$

$213,231$

$312,321$

for $\{1,2,3,4\}$ set with $4$ elements we have $$24=1\cdot2\cdot3\cdot4=4!$$ permutations

$1234,1243,1324,1342,1423,1432$

$2134,2143,2314,2341,2413,2431$

$3124,3142,3214,3241,3412,3421$

$4123,4132,4213,4231,41312,4321$

Continuing this way we have that set with n elements has $$1\cdot 2\cdot3\cdot...\cdot n=n!$$ permutations

Answer 3

So you mean ordered combinations that use all the elements in every combination. Since a set can't contain the same element more than once, you're talking about the number of permutations of the set, which is n!, where n is the size of the set.

Answer 4

The classroom thinking is : I have n boxes

How many elements can i can choose to put in the first box : $n$

How many elements can i can choose to put in the second box : $n - 1$ since i have put one in the first box.

And so on, so the total posibilities are $n \times n-1 \times \dots 1 = n!$, the factorial number.

Answer 5

$n! = n \cdot (n-1) \dots 1$

In C/C++:

int max = n ; //( 3 or 2 .. )
total = max;

while (max > 0){
     max --;
     total = total * max;
 }